Consider a completely full cylindrical water tank of height $1.6\ \text{m}$ and cross-sectional area $0.5\ \text{m}^2$. It has a small hole in its side at a height $90\ \text{cm}$ from the bottom. Assume the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load of $50\ \text{kg}$ is applied on the top surface of the water in the tank, then the velocity of the water coming out at the instant when the hole is opened is: ($g = 10\ \text{m/s}^2$)
Answer: (D) $4\ \text{m/s}$
The load adds a pressure $\dfrac{mg}{A} = \dfrac{500}{0.5} = 1000\ \text{Pa}$ on the top surface. The hole is $1.6 - 0.9 = 0.7\ \text{m}$ below the surface.
Bernoulli's equation from the top surface (speed negligible, since the tank is wide) to the hole (both open to the atmosphere apart from the load):
$$P_0 + \frac{mg}{A} + \rho g h = P_0 + \tfrac12\rho v^2$$
$$1000 + 1000\times10\times0.7 = \tfrac12\times1000\,v^2 \Rightarrow 8000 = 500\,v^2 \Rightarrow v = 4\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics